Question: Write the equation of a line that is perpendicular to $y=3x-2$ and that passes through the point $(-9,5)$.
Explanation: Getting started Key idea: The slopes of perpendicular lines are negative reciprocals of each other. Step 1: Find the slope Slope of the given line: ${3}$ So, the slope of the perpendicular line: $C{-\dfrac{1}{3}}$ Step 2: Substitute the known point into linear equation The perpendicular line will have a slope of $C{-\dfrac{1}{3}}$ and pass through the point ${(-9,5)}$. Let's start from the point-slope form of the equation of the perpendicular line, then solve for $y$. [What is the point-slope form?] $\begin{aligned} y-{5} &= C{-\dfrac{1}{3}}(x-{(-9)})\\\\\\ y-5 &= C{-\dfrac{1}{3}}x -3 \\\\\\ y &= C{-\dfrac{1}{3}}x { +2} \end{aligned}$ Answer $y=C{-\dfrac{1}{3}}x {+2}$. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ $y$ $x$